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The complexes that have a magnetic momen...

The complexes that have a magnetic moment of 1.73 BM is

A

`[Ti(H_(2)O)_(6)]^(3+)`

B

`[V(H_(2)O)_(6)]^(4+)`

C

`[Mn(H_(2)O)_(6)]^(2+)`

D

`[Mn(H_(2)O)_(6)]^(3+)`

Text Solution

Verified by Experts

The correct Answer is:
a,b

A & B: `3d^(1)` configuration.
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Which one of the following species doesn’t have a magnetic moment of 1.73 BM, (spin only value ) ?

Which of the following do not have magnetic moment of 1.73BM

Knowledge Check

  • Which of the complexes has the magnetic moment of 3.87 BM ?

    A
    `[Co(NH_(3))_(6)]^(3+)`
    B
    `[CoF_(6)]^(3-)`
    C
    `[CoCl_(4)]^(2-)`
    D
    `[Co(dmg)_(2)]` square planar complex (dmg= dimethyl glyoxime )
  • Among the following complexes which has magnetic moment of 5.9 BM

    A
    `Ni(CO)_(4)`
    B
    `[Fe(H_(2)O)_(6)]^(2+)`
    C
    `[Co(NH_(3))_(6)]^(2+)`
    D
    `[MnBr_(4)]^(2-)`
  • A compound of vanadium has a magnetic moment of 1.73 BM . What will be the electronic configurations:

    A
    `1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(6) 3d^(1)`
    B
    `1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(6) 3d^(2)`
    C
    `1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(6) 3d^(3)`
    D
    `1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(6) 3d^(4)`
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    A compound of vanadium has a magnetic moment of 1.73BM Work out the electronic configuration of vanadius in the compound

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    A compound of vanadium has a magnetic moment of 1.73 BM. The electronic configuration of vanadium ion in the compound is:

    Assertion (A) : Spin only formula to determine the magnetic movement of a substance is : mu=sqrt(n(n+2)) Reason (R) : A simple unpaired election has a magnetic moment of 1.73 BM.